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    <p>You have to use consistent types for objects declared across different translation units. </p> <p>Given <code>int a[] = {2, 3};</code>, either of the declarations <code>extern int a[];</code> or <code>extern int a[2]</code>; would be compatible whereas <code>extern int *a;</code> would not as pointers and arrays are completely separate types.</p> <p>The one thing special about arrays is the when the name of an array appears in any expression context other than as an operand to "address of" (unary <code>&amp;</code>) or <code>sizeof</code>, they are automatically converted to a pointer to their first element. This is what provides the syntax compatibility between arrays and pointers, but they are not the same type.</p> <p>Consider these two functions for a commented example. Note that the expression <code>a</code> is converted to a pointer to its first element in the second (and technically third) <code>printf</code> of the first function, but not in the first <code>printf</code> where it is the operand to <code>&amp;</code>.</p> <pre><code>#include &lt;stdio.h&gt; void print_array_info(void) { extern int a[]; printf("address of a: %p\n", (void*) &amp;a); // prints address of a printf(" converted a: %p\n", (void*) a); // prints address of a[0] printf("value of a[0]: %x\n", a[0]); // prints value of a } void print_pointer_info(void) { extern int a[]; int *b = a; // == &amp;a[0] printf("address of b: %p\n", (void*) &amp;b); // prints address of b printf(" value of b: %p\n", (void*) b); // prints value of b (== &amp;a[0]) printf("value of b[0]: %x\n", b[0]); // prints value of b[0] (== a[0]) } </code></pre> <p>Note that I use <code>%p</code> to print pointers and explicitly cast to <code>void*</code>.</p>
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